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I have a chi squared of 0.1185 with 1 degree of freedom. How can i determine the P-value?

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1  
Because $\chi^2(1)$ is the distribution of the square of a standard normal variate, your question is tantamount to asking for the probability that a standard normal value exceeds $\sqrt{0.1185}\approx0.34$ in size. That's an elementary calculation (and I'm sure you have software or accurate tables to carry it out). – whuber♦ Mar 5 '12 at 23:32

closed as not a real question by whuber♦ Aug 14 '12 at 12:38

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2 Answers

up vote 4 down vote

By p-value I presume you mean "the probability of seeing a value of my test statistic as large as this, if it really comes from a $\chi^2$ distribution with one degree of freedom". So if you get a low p-value it means it is an improbable value to have seen.

In R:

1-pchisq(0.1185, 1)

or

pchisq(0.1185, 1, lower=FALSE)

In Excel

=1-CHISQ.DIST(0.1185,1, TRUE)

or

=CHISQ.DIST.RT(A2,1)

The p-value is very large because your value of the test statistic is actually considerably less than the average value of such a $\chi^2$ random variable. If anything, it is unusually small. So basically, you do not have a large value and hence there is not evidence to dismiss the hypothesis that it genuinely comes from a $\chi^2$ distribution.

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so, what is the p-value? – user176105 Mar 5 '12 at 22:54
I would recommend you always use FALSE instead of F, so it should be pchisq(0.1185, 1, lower = FALSE). – Henrik Mar 5 '12 at 23:08
@user176105 I'll leave it up to you to calculate... you must have access to either R or Excel, surely. – Peter Ellis Mar 5 '12 at 23:10
@PeterEllis so my teacher just gave me a table like this: sociology.ohio-state.edu/people/ptv/publications/p%20values/… so the only thing i can do is narrow my p-value to a range? – user176105 Mar 5 '12 at 23:14
3  
Linear interpolation of the 2-tailed p-value for $z=\sqrt{0.1185}$ gives $0.73$ which is fully accurate to two d.p. (In R this can be done as (sqrt(0.1185)-.39)/(.25-.39) * (.80-.70) + 0.70; the numbers 0.39, 0.25, 0.80, and 0.70 come from your teacher's table.) Alternatively, applying Simpson's Rule (x<-0.1185;1 - sqrt(x)*(2*exp(-x/2) + 4)/(3*sqrt(2*pi))) gives 0.7306, which is within 0.0001 of the truth: no table is needed at all. – whuber♦ Mar 5 '12 at 23:43
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up vote 1 down vote

The upper tail of the $\chi^2_1$ distribution, beyond 0.1185 is

pchisq(0.1185, df=1, lower=F)
[1] 0.7306671

That's your p-value

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How did you get that exact number? my teacher gave me a chart and I could tell my value was between 0.7 and 0.9, but I couldn't get any more accurate – user176105 Mar 5 '12 at 22:41
@Henrik Why use FALSE instead of F? F seems to work – Peter Flom Mar 5 '12 at 23:20
@guest was using R. – Peter Flom Mar 5 '12 at 23:20
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In fairness to @guest, the homework tab didn't appear till after s/he'd answered the question – Peter Ellis Mar 6 '12 at 0:25
3  
@PeterFlom Because only TRUE and FALSE are reserved keywords (see ?reserved). F and T are just variables assigned to your convenience. So you can easily use the following construction to produce horrible things: T <- FALSE; F <- TRUE. – Henrik Mar 6 '12 at 7:41
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